Math 211 Fall 2007: Solutions: HW #3
Instructor: S. Cautis
1. section 2.6, #2
dF = (2x − y)dx + (−x + 2y)dy
2. section 2.6, #16
2u
(2u)(2v)
∂
2v
∂
(
)= 2
=
(
)
∂v u2 + v 2
(u + v 2 )
∂u u2 + v 2
so the equation is exact!
Solving
∂
∂u F
=
2u
u2 +v 2
by integrating with respect to u we get
F (u, v) = ln(u2 + v 2 ) + φ(v)
Taking the derivative with respect to v we must get
2v
2v
+ φ0 (v) = 2
u2 + v 2
u + v2
so φ(v) = C is a constant. Thus F = ln(u2 + v 2 ) + C.
3. section 2.6, #24
Integrating
3(y+1)2
x4
with respect to x we get
(y + 1)2
+ φ(y)
x3
Taking the derivative with respect to y we get
F (x, y) = −
−
2(y + 1)
2x(y + 1)
+ φ0 (y) = −
x3
x4
2
from which we get φ0 (y) = 0 so φ(y) = C constant. Thus F = − (y+1)
x3 +C.
4. section 2.6, #28
We want µ(y) such that
∂
∂y (2yµ)
=
∂
∂x ((x
+ y)µ) or equivalently
2µ + 2yµ0 = µ
√
Thus 2yµ0 = −µ. Solving by separating variables we get µ = 1/ y. Thus
we need to solve
√
√
2 ydx + (x + y)/ ydy = 0
Integrating the first term with respect to x we get
√
F = 2x y + φ(y)
and taking the derivative with respect to y gives
√
√
x/ y + φ0 = (x + y)/ y
√
or φ0 = y which means φ = 2/3y 3/2 + C.
5. section 2.6, #36
A quick check shows it is not exact. Fortunately, it is homogeneous so we
can make the substitution y = xv to get
(x + xv)dx + (xv − x)(xdv + vdx) = 0
or equivalently
(1 + v)dx + (v − 1)(xdv + vdx) = 0
or
(1 + v 2 )dx + x(v − 1)dv = 0
which is separable. We separate to get
dx/x = −(v − 1)/(1 + v 2 )dv
so integrating we get
ln |x| = −1/2 ln(1 + v 2 ) + arctan(v) + C
Substituting back v = y/x we get the general solution
− ln |x| − 1/2 ln(1 + (y/x)2 ) + arctan(y/x) + C = 0
6. section 2.7, #1
f (t, y) = 4 + y 2 is continuous everywhere and so is
unique solution is guaranteed.
∂
∂y f (t, y)
= 2y so a
7. section 2.7, #2
√
∂
1
f (t, y) = y is continuous everywhere and ∂y
f (t, y) = 2√
y is continous
and well defined for y > 0. Our initial condition is at t0 = 4 and y0 = 0 and
so there is no rectangle containing this point where both are continuous –
so hypothesis of theorem is not satisfied.
8. section 2.7 #6
∂
f (x, y) = −y/x2 is also
f (x, y) = y/x + 2 is continuous for x 6= 0 while ∂x
continuous for x 6= 0. Since initial condition is (x0 , y0 ) = (0, 1) there is no
rectangle again and so theorem does not guarantee a unique solution.
9. section 2.7 #8
The equation is linear. The general solution is y(t) = t + 2Ct2 . All of
these solutions pass through the point (0, 0) and look like (are in fact)
a parabola. This means that y(0) = 2 has no solution. If we put this
equation into normal form we get
dy
= 2y/t − 1
dt
which is not continuous at t = 0. Consequently the hypothesis of the
existence theorem are not satisfied.
10. section 2.7 #21
(a) We have
y 0 (t+
0) =
=
lim+
t→t0
(t − t0 )3 − 0
t − t0
lim (t − t0 )2
t→t+
0
= 0
Similarly,
y 0 (t−
0) =
lim
t→t−
0
= 0
0−0
t − t0
So the two limits agree and thus y(t) is differentiable at t = t0 with
y 0 (t0 ) = 0 since both right and left hand derivatives are equal to 0.
(b) In this case f (t, y) = 3y 2/3 is continuous at (0, 0) (so solution exists
∂
by theorem) but ∂y
f (t, y) = 2y −1/3 which is not continuous at (0, 0).
Hence theorem 7.16 does not guarantee a unique solution there since the
hypothesis is not satisfied (and indeed, as we see, there are at least two
solutions).